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The engineer of a passenger train traveling at 25.0m/s sights a freight train whose caboose is 200m ahead on the same track. The freight train is traveling at 15.0m/s in the same direction as the passenger train. The engineer of the passenger train immediately applies the brakes, causing a constant accelaration of -0.100m/s2, while the freight train continues with constant speed. Take x=0 at the location of the front of the passenger train when the engineer applies the brake.
a) Will the cows nearby witness a collision?
b) If so, where does it take place
c) On a single graph, sketch the positions of the frong of the passenger train and the back of the freight train.

Sagot :

Answer:

a) if the two trains are going to collide

b) x = 538 m

Explanation:

To solve this exercise we use the kinematics relations in one dimension.

We set a reference system at the starting point of the passenger train

Freight train, we use index 1 for this train

          x₁ = x₀ + v₁ t

passenger train

          x₂ = v₂ t - ½ a₂ t²

as we are using the same system to measure the position of the two trains, at the meeting point the position must be the same

           x₁ = x₂

we substitute

          x₀ + v₁ t = v₂ t - ½ a₂ t²

         ½ a₂ t² + t (v₁ -v₂) + x₀ = 0

we subjugate the values

         ½ 0.1 t² + t (15-25) + 200 = 0

         0.05 t² - 10 t + 200 = 0

          t² - 200 t + 4000 = 0

we solve the second degree system

         t = [200 ±[tex]\sqrt{200^2 - 4 \ 4000}[/tex] ]/ 2

         t = [200 ± 154.9] / 2

         t₁ = 177.45 s

         t₂ = 22.55 s

therefore for the smallest time  the two trains must meet  t₂ = 22.55 s

merchandise train

            x = 200+ 15  22.55

            x = 538 m

passenger train

           x = 25 22.55 -1/2  0.100  22.55²

           x = 538 m

we see that the trains meet for this distance

a) if the two trains are going to collide

b) x = 538 m

c) see attachment for schematic graphics

View image Moya1316
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